The Double Angle Formulas

Put B = A: sin 2A = 2 sin A cos A

The compound angle formula sin(A + B) = sin A cos B + cos A sin B holds for every A and B, so it holds when B is equal to A. Then A + B is A + A = 2A, and the formula becomes sin 2A = sin A cos A + cos A sin A.

The two products are the same, so they add to twice one of them: sin 2A = 2 sin A cos A. This is the double angle formula for sine.

Check it at A = 30°: $2 \sin 30^\circ \cos 30^\circ = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$, which is sin 60°. In radians, at A = 0.3, 2 sin 0.3 cos 0.3 = 2 × 0.2955 × 0.9553 = 0.5646, which is sin 0.6; at A = 1.1, 2 × 0.8912 × 0.4536 = 0.8085, which is sin 2.2.

Doubling an angle does not double its sine. 2 sin 30° = 1, but sin 60° is $\frac{\sqrt{3}}{2}$, about 0.87. The cos A in the formula is what corrects the doubling.

The gold curve is y = sin 2x and the plain curve is y = 2 sin x, over one turn with one square across for every 90°. Doubling the angle squeezes the wave into half the width and leaves its height at 1; doubling the sine stretches it to a height of 2. The dots at 45° show the difference: sin 90° = 1, but $2 \sin 45^\circ = \sqrt{2}$, about 1.41.

The same move in cosine

Put B = A in cos(A + B) = cos A cos B − sin A sin B. It becomes cos 2A = cos A cos A − sin A sin A, which is $\cos 2A = \cos^{2}A - \sin^{2}A$.

Check it at A = 30°: $\cos^{2}30^\circ - \sin^{2}30^\circ = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$, which is cos 60°. In radians, $\cos^{2}0.3 - \sin^{2}0.3 = 0.8253$ to 4 decimal places, which is cos 0.6, and $\cos^{2}1.1 - \sin^{2}1.1 = -0.5885$, which is cos 2.2. The answer is negative there because 2.2 radians is more than a quarter turn.

Three forms of cos 2A

The identity $\sin^{2}A + \cos^{2}A = 1$ lets either square be replaced by the other. Replace $\cos^{2}A$ with $1 - \sin^{2}A$: $\cos 2A = \left(1 - \sin^{2}A\right) - \sin^{2}A = 1 - 2\sin^{2}A$.

Replace $\sin^{2}A$ with $1 - \cos^{2}A$ instead: $\cos 2A = \cos^{2}A - \left(1 - \cos^{2}A\right) = 2\cos^{2}A - 1$.

So there are three forms: $\cos 2A = \cos^{2}A - \sin^{2}A = 1 - 2\sin^{2}A = 2\cos^{2}A - 1$. At A = 30°, all three give ½: $\frac{3}{4} - \frac{1}{4}$, then $1 - 2 \times \frac{1}{4}$, then $2 \times \frac{3}{4} - 1$. Choose the form whose square matches the other terms of the problem, so that only one function is left.

Turned around: $\sin^{2}A$ and $\cos^{2}A$

Start from $\cos 2A = 1 - 2\sin^{2}A$. Add $2\sin^{2}A$ to both sides and subtract cos 2A: $2\sin^{2}A = 1 - \cos 2A$. Divide by 2: $\sin^{2}A = \frac{{1 - \cos 2A}}{2}$. Here the equation is being rearranged to make $\sin^{2}A$ the subject, not proved, so working on both sides is allowed.

The same steps on $\cos 2A = 2\cos^{2}A - 1$ give $\cos^{2}A = \frac{{1 + \cos 2A}}{2}$. The minus sign goes with $\sin^{2}A$ and the plus sign with $\cos^{2}A$.

Check at 1.1 radians: $\frac{{1 - \cos 2.2}}{2} = \frac{{1 + 0.5885}}{2} = 0.7943$, which is $\sin^{2}1.1$. These forms turn a square of a trigonometric function into a plain cosine, and that is how $\sin^{2}x$ and $\cos^{2}x$ are integrated.

The gold curve is $y = \sin^{2}x$ and the plain curve is y = cos 2x, over one turn with one square across for every 90°; the plain straight line is y = ½. The gold wave is the plain wave turned upside down, halved in height and raised to the level ½: where cos 2x is 1, $\sin^{2}x$ is 0, and where cos 2x is −1, $\sin^{2}x$ is 1. That is $\sin^{2}x = \frac{1}{2} - \frac{1}{2} \cos 2x$.

Using the formulas

Suppose A is acute and $\sin A = \frac{3}{5}$. Then $\cos^{2}A = 1 - \frac{9}{25} = \frac{16}{25}$, and $\cos A = \frac{4}{5}$ because A is acute. So $\sin 2A = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$ and $\cos 2A = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$. Check: $\left(\frac{24}{25}\right)^{2} + \left(\frac{7}{25}\right)^{2} = \frac{{576 + 49}}{625} = 1$. On a calculator, A is 36.87°, and sin 73.74° = 0.96 and cos 73.74° = 0.28, which are $\frac{24}{25}$ and $\frac{7}{25}$.

The $\cos^{2}A$ form gives exact values for half of a known angle. With A = 15°, $\cos^{2}15^\circ = \frac{{1 + \cos 30^\circ}}{2} = \frac{{1 + \frac{\sqrt{3}}{2}}}{2} = \frac{{2 + \sqrt{3}}}{4}$. So $\cos 15^\circ = \frac{\sqrt{2 + \sqrt{3}}}{2}$, which is 0.9659 to 4 decimal places, the same as a calculator gives.

Solving an equation with a double angle

Solve sin 2x = sin x for $0^\circ \le x \le 360^\circ$. The equation mixes the angles 2x and x, so first rewrite sin 2x: 2 sin x cos x = sin x.

Bring every term to one side and factor: 2 sin x cos x − sin x = 0, so sin x (2 cos x − 1) = 0. Either sin x = 0, which gives x = 0°, 180° or 360°, or cos x = ½, which gives x = 60° or 300°.

Do not divide both sides by sin x: that throws away the three solutions where sin x = 0. Check x = 60°: $\sin 120^\circ = \frac{\sqrt{3}}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Check x = 300°: $\sin 600^\circ = \sin 240^\circ = - \frac{\sqrt{3}}{2}$, and $\sin 300^\circ = - \frac{\sqrt{3}}{2}$.

The usual mistakes

Writing sin 2A = 2 sin A. At A = 30° that gives 1, but sin 60° is $\frac{\sqrt{3}}{2}$. The formula is 2 sin A cos A.

Writing $\cos 2A = \cos^{2}A + \sin^{2}A$. That sum is the Pythagorean identity, and it equals 1, not cos 2A; the cosine formula has a minus sign.

Mixing up the signs when rearranging. $\sin^{2}A = \frac{{1 - \cos 2A}}{2}$ and $\cos^{2}A = \frac{{1 + \cos 2A}}{2}$. Check with A = 0: $\sin^{2}0 = 0$ and cos 0 = 1, so the sine form needs the minus.

Forgetting to halve. 1 − cos 2A is $2\sin^{2}A$, so $\sin^{2}A$ is half of it.

The launch angle for the greatest range

In the application below, a ball thrown at $20 \frac{m}{s}$ at an angle $\theta$ lands $80 \sin \theta \cos \theta m$ away. The double angle formula turns that into $40 \sin 2\theta$, and a sine is greatest, 1, when $2\theta$ is a right angle, so the best angle is 45°.